## PROBLEM TO BE SOLVED

Let’s say, there is a huge array ‘A’ of size N, N = 1000000.

Problem (query) is to find the sum of all the numbers in A from 100 till 2000 i.e. A[100] + A[101] + A[102] + A[103] + ….. + A[1998] + A[1999] + A[2000].

Well, a simple way is this

int sum = 0;
for(int i=100; i<=2000; i++) sum += A[i];
print(sum)

This is O(n)

Let’s make these modifications on A:

A[999] = 23, A[2] = 23, A[3412] = 23, A[3] = 23, .. etc.

After these modifications we want to calculate the sum from A[100] till A[2000]. How will we do this?

For every single query, we have to loop from 100 to 2000 again and again. So if there are ‘m’ queries then the run time is O(m*n).

Problem becomes even more difficult if we need to calculate the sum from any index ‘i’ to index ‘j’ of A ,(i<j), i.e.

A[i] + A[i+1] + A[i+2] + ... + A[j-1] + A[j]

Even if we create a new array Sum such that

Sum[ind] = Sum[ind-1] + A[ind]

we have to update this new array for every change made to A.

Can we do better? Can we avoid this O(n) for every change to A. Let’s see.

## WHAT is “num & (-num)” ? Here ‘&’ is Logical AND

Say there is a integer num = 5. Binary representation of ‘5’ is ‘101’. But what is the binary representation of ‘-5’ ?

Number |
Binary representation |

5 |
101 |

-5 |
? |

First, let’s understand how negative of a number is created.

Negative of NUM = ( Complement of the Num + 1)
Complement of the Num = Convert all 1's to 0 and 0's to 1

Let's calculate Binary of '-5'
Binary of '5' = '101'
So, Complement of 5 = Complement of '101' = '010' [convert 1 to 0 and 0 to 1]
So, as per rule above, Negative of 5 = ('010' + 1) = '011'
So, the binary form of '-5' = '011'

Num |
Num in Binary |
Binary representation of (-Num) |

5 |
101 |
011 |

7 |
111 |
001 |

8 |
100 |
100 |

12 |
1100 |
0100 |

What is '5 & (-5)'?
'101' & '011' = '001'

Num, -Num |
Num & -Num |

101, 011 |
001 |

111,001 |
001 |

100,100 |
100 |

1100,0100 |
0100 |

As we can see all the “num & -num” result in a number with only the last ‘1’ in the original num. We will use this information soon.

## Going from ind to “ind + (ind & -ind)”

Let’s say ind = 1, Let’s go from ind to ind + (ind & -ind) till we reach ‘8’ or bigger. Try to do with you hand once.

Below table shows this:

ind (Binary) |
(ind & -ind) (Binary) |
ind + (ind & -ind) (Binary) |

1 (1) |
1(1) |
2(10) |

2(10) |
2(10) |
4(100) |

4(100) |
4(100) |
8(100) |

while(ind<=8) ind = ind + (ind & -ind);
So, Starting from ind = 1, ind changes like this: 1 -> 2 -> 4 -> 8
Starting from ind = 2, 2 -> 4 -> 8
Starting from ind = 3, 3 -> 4 -> 8
Starting from ind = 4, 4 -> 8
Starting from ind = 5, 5 -> 6 -> 8
Starting from ind = 6, 6 -> 8
Starting from ind = 7, 7 -> 8
Starting from int = 8, 8

while(ind >=0) ind = ind - (ind & -ind);
So, Starting from ind = 1, ind changes like this: 1 -> 0
Starting from ind = 2, 2 -> 0
Starting from ind = 3, 3 -> 2 -> 0
Starting from ind = 4, 4 -> 0
Starting from ind = 5, 5 -> 4 -> 0
Starting from ind = 6, 6 -> 4 -> 0
Starting from ind = 7, 7 -> 5 -> 4 -> 0
Starting from int = 8, 8 -> 0

I recommend solving both above code by hand once.

## Binary Indexed Tree (BIT)

Let's say, our array A is of size N=8, A[0...7]
We create a new array called BIT[0..8]
0 1 2 3 4 5 6 7 8
BIT[] = {0, 0, 0, 0, 0, 0, 0, 0, 0}
A[] = {0, 0, 0, 0, 0, 0, 0, 0, 0}

Now let’s run the following code:

A[0] = 3
updateBIT(0, A[0])
void updateBIT(int ind, int val){
ind = ind + 1; // Increase ind by 1
while(ind <= N){
// increase the value of BIT[ind] by val
BIT[ind] = BIT[ind] + val;
// go to next of ind
ind = ind + (ind & -ind);
}
}

Note that: updateBIT first change ind to ind+1. What is gonna happen?

We know, for ind = 1, ind changes like this: 1 -> 2 -> 4 -> 8
So, at the end of the above code BIT will look like this:
0 1 2 3 4 5 6 7 8
BIT[] = {0, 3, 3, 0, 3, 0, 0, 0, 3}
A[] = {0, 3, 0, 0, 0, 0, 0, 0, 0}

Now let’s run the following code:

A[1] = 2
updateBIT(1, A[1])

What is gonna happen? Again, ind = ind+1 before the while loop.

We know, for ind = 2, ind changes like this: 2 -> 4 -> 8
So, at the end of the above code BIT will look like this:
0 1 2 3 4 5 6 7 8
BIT[] = {0, 3, 5, 0, 5, 0, 0, 0, 5}
A[] = {0, 3, 2, 0, 0, 0, 0, 0, 0}

If we keep doing like this till ind=7. We will see something like this:

BIT[1] = A[0]
BIT[2] = A[0] + A[1]
BIT[3] = A[2]
BIT[4] = A[0] + A[1] + A[2] + A[3]
BIT[5] = A[4]
BIT[6] = A[4] + A[5]
BIT[7] = A[6]
BIT[8] = A[0] + A[1] + A[2] + A[3] + A[4] + A[5] + A[6] + A[7]

## Querying on BIT

How shall we answer the query: Get the sum from A[2] till A[5] i.e.

A[2] + A[3] + A[4] + A[5] =??

Let’s define this function

int getSum(int ind){
ind = ind + 1;
int sum = 0;
while(ind > 0){
sum = sum + BIT[ind];
ind = ind - (ind & -ind);
}
return sum;
}

Let’s run getSum with different ind.

ind = 0, sum = BIT[1]
ind = 1, sum = BIT[2], from above table we know BIT[2] = A[0] + A[1]
ind = 2, sum = BIT[3] + BIT[2] => A[2] + A[1] + A[0]
etc.

Note that here we are going in opposite direction because of ‘-‘ in

ind = ind **-** (ind & -ind);

getSum(5) will run like this:
sum = BIT[6] + BIT[4]
From above tables, sum = (A[4] + A[5]) + (A[0] + A[1] + A[2] + A[3])
So getSum(1) will run like this:
sum = BIT[2]
From above tables, sum = (A[0] + A[1])
=> getSum(5) - getSum(1) = A[2] + A[3] + A[4] + A[5]
So to get sum from A[2] till A[5], we need
getSum(5) - getSum(1)
In other words to get sum from A[i] till A[j], we need to calculate
getSum(j) - getSum(1-1)

That’s how we query BIT.

## Runtime of updateBIT and getSum

For both updateBIT and getSum,average runtime is O(Log N). This is because maximum bits in a binary representation of a number N can be Log N.

Till Next time. CHAO!